Lemma 3

Let $A$ be a commutative algebra of finite degree over the algebraically closed field $K$, such that $\Omega_K(A) = 0$. Then we have $m = m^2$ for every maximal ideal $m$ of $A$.

The algebra $A/m$ is an extension of finite degree of the algebraically closed field $K$, whence $[A/m : K] = 1$. Hence for each $a \in A$ there exists a unique scalar $\lambda$ such that $a - \lambda \cdot 1 \in m$; write $D(a)$ for the residue class of $a - \lambda \cdot 1$ mod $m^2$. It is clear that $D$ is a $K$-derivation of $A$ into the $A$-module $m/m^2$. The universal property of $\Omega_K(A)$ (III, p. 569) and the hypothesis $\Omega_K(A) = 0$ now imply that $D = 0$, whence $m/m^2 = 0$ and so $m = m^2$.